Double Integral with Negative Exponent

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SUMMARY

The discussion centers on evaluating the double integral ∫4to5 ∫1to2 (1x + y)−2 dy dx. Participants clarify that the negative exponent indicates the need to rewrite the integrand as (x + y)−2, which simplifies the integration process. An antiderivative of -2 with respect to y is also mentioned, emphasizing the importance of correctly interpreting the exponent in the context of integration.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with negative exponents and their implications
  • Knowledge of substitution methods in integration
  • Basic skills in evaluating antiderivatives
NEXT STEPS
  • Study techniques for evaluating double integrals with variable limits
  • Learn about substitution methods in integration, specifically for negative exponents
  • Explore the properties of antiderivatives and their applications in calculus
  • Practice problems involving integrals with negative exponents
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Students studying calculus, particularly those focusing on integration techniques, and educators looking for examples of double integrals with negative exponents.

Loppyfoot
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Homework Statement



∫4to5 ∫1to2 (1x + y)−2 dy dx





The Attempt at a Solution



I am confused about what to do with this negative 2.

Any ideas?
 

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Loppyfoot said:

Homework Statement



∫4to5 ∫1to2 (1x + y)−2 dy dx





The Attempt at a Solution



I am confused about what to do with this negative 2.

Any ideas?

An antiderivative of -2 with respect to y is -2y. Or if that -2 is supposed to be an exponent, use the X2 button: (x + y)-2. If that is what you mean then let

u = x + y, du = dy

to proceed.
 

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